# Check if a given directed graph is strongly connected | Set 2 (Kosaraju using BFS)

Given a directed graph, find out whether the graph is strongly connected or not. A directed graph is strongly connected if there is a path between any two pairs of vertices. There are different methods to check the connectivity of directed graph but one of the optimized method is Kosaraju’s DFS based simple algorithm.
Kosaraju’s BFS based simple algorithm also work on the same principle as DFS based algorithm does.

```Following is Kosaraju’s BFS based simple algorithm
that does two BFS traversals of graph:
1) Initialize all vertices as not visited.

2) Do a BFS traversal of graph starting from
any arbitrary vertex v. If BFS traversal
doesn’t visit all vertices, then return false.

3) Reverse all edges (or find transpose or reverse
of graph)

4) Mark all vertices as not visited in reversed graph.

5) Again do a BFS traversal of reversed graph starting
from same vertex v (Same as step 2). If BFS traversal
doesn’t visit all vertices, then return false.
Otherwise, return true.
```

The idea is again simple if every node can be reached from a vertex v, and every node can reach same vertex v, then the graph is strongly connected. In step 2, we check if all vertices are reachable from v. In step 5, we check if all vertices can reach v (In reversed graph, if all vertices are reachable from v, then all vertices can reach v in original graph).

Explanation with some examples:

Example 1 :

Given a directed to check if it is strongly connected or not.

step 1: Starting with vertex 2 BFS obtained is 2 3 4 0 1

step 2: After reversing the given graph we got listed graph.

step 3: Again after starting with vertex 2 the BFS is 2 1 4 0 3

step 4: No vertex in both case (step 1 & step 3) remains unvisited.

step 5: So, given graph is strongly connected.

Example 2 :

Given a directed to check if it is strongly connected or not.

step 1: Starting with vertex 2 BFS obtained is 2 3 4

step 2: After reversing the given graph we got listed graph.

step 3: Again after starting with vertex 2 the BFS is 2 1 0

step 4: vertex 0, 1 in original graph and 3, 4 in reverse graph remains unvisited.

step 5: So, given graph is not strongly connected.

Following is the implementation of above algorithm.

```// C++ program to check if a given directed graph
// is strongly connected or not with BFS use
#include <bits/stdc++.h>
using namespace std;

class Graph
{
int V;    // No. of vertices

// A recursive function to print DFS starting from v
void BFSUtil(int v, bool visited[]);
public:

// Constructor and Destructor
Graph(int V) { this->V = V;  adj = new list<int>[V];}
~Graph() { delete [] adj; }

// Method to add an edge

// The main function that returns true if the
// graph is strongly connected, otherwise false
bool isSC();

// Function that returns reverse (or transpose)
// of this graph
Graph getTranspose();
};

// A recursive function to print DFS starting from v
void Graph::BFSUtil(int v, bool visited[])
{
// Create a queue for BFS
list<int> queue;

// Mark the current node as visited and enqueue it
visited[v] = true;
queue.push_back(v);

// 'i' will be used to get all adjacent vertices
// of a vertex
list<int>::iterator i;

while (!queue.empty())
{
// Dequeue a vertex from queue
v = queue.front();
queue.pop_front();

// Get all adjacent vertices of the dequeued vertex s
// If a adjacent has not been visited, then mark it
// visited and enqueue it
{
if (!visited[*i])
{
visited[*i] = true;
queue.push_back(*i);
}
}
}
}

// Function that returns reverse (or transpose) of this graph
Graph Graph::getTranspose()
{
Graph g(V);
for (int v = 0; v < V; v++)
{
// Recur for all the vertices adjacent to this vertex
list<int>::iterator i;
}
return g;
}

{
}

// The main function that returns true if graph
// is strongly connected
bool Graph::isSC()
{
// St1p 1: Mark all the vertices as not
// visited (For first BFS)
bool visited[V];
for (int i = 0; i < V; i++)
visited[i] = false;

// Step 2: Do BFS traversal starting
// from first vertex.
BFSUtil(0, visited);

// If BFS traversal doesn’t visit all
// vertices, then return false.
for (int i = 0; i < V; i++)
if (visited[i] == false)
return false;

// Step 3: Create a reversed graph
Graph gr = getTranspose();

// Step 4: Mark all the vertices as not
// visited (For second BFS)
for(int i = 0; i < V; i++)
visited[i] = false;

// Step 5: Do BFS for reversed graph
// starting from first vertex.
// Staring Vertex must be same starting
// point of first DFS
gr.BFSUtil(0, visited);

// If all vertices are not visited in
// second DFS, then return false
for (int i = 0; i < V; i++)
if (visited[i] == false)
return false;

return true;
}

// Driver program to test above functions
int main()
{
// Create graphs given in the above diagrams
Graph g1(5);
g1.isSC()? cout << "Yes\n" : cout << "No\n";

Graph g2(4);
g2.isSC()? cout << "Yes\n" : cout << "No\n";

return 0;
}
```

Output:

```Yes
No
```

Time Complexity: Time complexity of above implementation is same as Breadth First Search which is O(V+E) if the graph is represented using adjacency matrix representation.

Can we improve further?
The above approach requires two traversals of graph. We can find whether a graph is strongly connected or not in one traversal using Tarjan’s Algorithm to find Strongly Connected Components.

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